# 2134. Minimum Swaps to Group All 1's Together II

cicular array just append same array on it

but sliding win size (one count), should cal from the original array

T : O(n)

S: O(n)

```java
class Solution {
    public int minSwaps(int[] nums) {
        int n = nums.length;

        int winSize = 0; // all 1's togeter will be the window size
        
        // for circular array, just need " append another same array " on the original array
        int[] data = new int[2*n]; 
        for (int i = 0; i < n; i++) {
            if (nums[i] == 1) {
                winSize++;
            }
            data[i] = nums[i];
            data[i+n] = nums[i];
        }
        
        int left = 0;
        int oneCount = 0; // count in this window, how many one
        int res = winSize;
        
        for (int right = 0; right < data.length; right++) {
            if (right - left + 1 > winSize) {
                oneCount -= data[left++];
            }
            oneCount += data[right];
            
            if (right - left + 1 == winSize) {
                // 0 needs to swap = winSize(total of ones) - one's count in window
                res = Math.min(res, winSize - oneCount); 
            }
        }
        return res;
        
    }
}
```

```java
class Solution {
    public int minSwaps(int[] nums) {
        int n = nums.length;
        int oneWinSize = 0;

        int[] newNum = new int[n*2];
        for (int i = 0; i < n; i++) {
            if (nums[i] == 1) {
                oneWinSize++;
            }
            newNum[i] = nums[i];
            newNum[i+n] = nums[i]; 
        }

        if (oneWinSize == 0 || oneWinSize == 1) {
            return 0;
        }

        int left = 0;
        int result = Integer.MAX_VALUE;
        int zeroCount = 0;
        for (int right = 0; right < n*2; right++) {
            if (newNum[right] == 0) {
                zeroCount++;
            }
            if (right - left + 1 > oneWinSize) {
                if (newNum[left] == 0) {
                    zeroCount--;
                }
                left++;
            }
            if (right - left + 1 == oneWinSize) {
                result = Math.min(result, zeroCount);
            }
        }
        return result;
    }
}

/**
find how many one first




 0 1 2 3 4 5 6 7
[0,1,1,1,0,0,1,1,0][0,1,1,1,0,0,1,1,0]

for ex: this has 5 one

so try to find a win with 5 size, the min zero in this won -> it's the ans 

6 one
[1,1,1,0,0,1,0,1,1,0][1,1,1,0,0,1,0,1,1,0]
               x x x  x x x
for this win [1,1,0][1,1,1] , only one zero in this window -> so ans is 1




Inital wrong thought:
if we only try to find a max win has 6 one, in the end max win - one's count
-> in  [1,1,1,0,0,1,0,1,1,0][1,1,1,0,0,1,0,1,1,0]
will be wrong
because a win has 6 one, is 1,0,1,1,0][1,1,1 or 1,1,1,0,0,1,0,1,1 (become 8 - 6 = 2 or 9 - 6 = 3)
both are not the correct ans
 */
```


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://timmybeeflin.gitbook.io/cracking-leetcode/two-pointer/sliding-window/2134.-minimum-swaps-to-group-all-1s-together-ii.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.